Motivation.

Exersize 6.1 from [1] is to show that the traction vector can be written in vector form (a rather curious thing to have to say) as

Note that the text uses a wedge symbol for the cross product, and I’ve switched to standard notation. I’ve done so because the use of a Geometric-Algebra wedge product also can be used to express this relationship, in which case we would write

In either case we have

(where the primes indicate the scope of the gradient, showing here that we are operating only on , and not ).

After computing this, lets also compute the stress tensor in cylindrical and spherical coordinates (a portion of that is also problem 6.10), something that this allows us to do fairly easily without having to deal with the second order terms that we encountered doing this by computing the difference of squared displacements.

We’ll work primarily with just the strain tensor portion of the traction vector expressions above, calculating

We’ll see that this gives us a nice way to interpret these tensor relationships. The interpretation was less clear when we computed this from the second order difference method, but here we see that we are just looking at the components of the force in each of the respective directions, dependent on which way our normal is specified.

Verifying the relationship.

Let’s start with the the plain old cross product version

We can also put the double cross product in wedge product form

Equivalently (and easier) we can just expand the dot product of the wedge and the vector using the relationship

We can now move on to compute the directional derivatives and complete the strain calculation in cylindrical coordinates. Let’s consider this computation of the stress for normals in each direction in term.

With .

Our directional derivative component for a normal direction doesn’t have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Our directional derivative component for a normal direction will have some cross terms since both and are functions of

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

With .

Like the normal direction, our directional derivative component for a normal direction will not have any cross terms

Projecting our curl bivector onto the direction we have

Putting things together we have

For our stress tensor

we can now read off our components by taking dot products to yield

\begin{subequations}

\end{subequations}

Summary.

\begin{subequations}

\end{subequations}

Spherical strain tensor.

Having done a first order cylindrical derivation of the strain tensor, let’s also do the spherical case for completeness. Would this have much utility in fluids? Perhaps for flow over a spherical barrier?

We need the gradient in spherical coordinates. Recall that our spherical coordinate velocity was

and our gradient mirrors this structure

We also previously calculated \inbookref{phy454:continuumL2}{eqn:continuumL2:1010} the unit vector differentials

\begin{subequations}

\end{subequations}

and can use those to read off the partials of all the unit vectors

Finally, our velocity in spherical coordinates is just

from which we can now compute the curl, and the directional derivative. Starting with the curl we have

So we have

With .

The directional derivative portion of our strain is

The other portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This is consistent with (15.20) from [3] (after adjusting for minor notational differences).

With .

Now let’s do the direction. The directional derivative portion of our strain will be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Putting these together we find

Which gives

For our stress tensor

we can now read off our components by taking dot products

\begin{subequations}

\end{subequations}

This again is consistent with (15.20) from [3].

With .

Finally, let’s do the direction. This directional derivative portion of our strain will also be a bit more work to compute because we have variation of the unit vectors

So we have

and can move on to projecting our curl bivector onto the direction. That portion of our strain tensor is

Sometimes we can utilize solutions already found to understand the behaviour of more complex systems. Combining the two we can look at flow over a plate as in figure (\ref{fig:continuumL12:continuumL12fig3})

Example 2. Fluid in a container. If the surface tension is altered on one side, we induce a flow on the surface, leading to a circulation flow. This can be done for example, by introducing a heat source or addition of surfactant.

This sort of flow is hard to analyze, only first done by Steve Davis in the 1980’s. The point here is that we can use some level of intuition to guide our attempts at solution.

Flow down a pipe.

Reading: section 2 from [1].

Recall that the Navier-Stokes equation is

We need to express this in cylindrical coordinates as in figure (\ref{fig:continuumL12:continuumL12fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL12fig5}
\caption{Flow through a pipe.}
\end{figure}

Our gradient is

For our Laplacian we find

which we can write as

NS takes the form

For steady state and incompressible fluids in the absence of body forces we have

or, in coordinates

With an assumption that we have no radial or circulatory flows (), and with assumed to only have a radial dependence, our velocity is

and an assumption of linear pressure dependence

then NS takes the final simple form

Solving this we have

Requiring finite solutions for means that we must have . Also , we have so we must have

and seek simultaneous solutions to the pair of stress tensor equations

In general this requires an iterated approach, solving for one with an initial approximation of the other, then switching and tuning the numerical method carefully for convergence.

We expect that the flow of liquid will induce a flow of air at the interface, but may be able to make a one-sided approximation. Let’s see how far we get before we have to introduce any approximations and compute the traction vector for the liquid

So

Our boundary value condition is therefore

When can we decouple this, treating only the liquid? Observe that we have

so if

we can treat only the liquid portion of the problem, with a boundary value condition

Let’s look at the component of the traction vector in the direction of the normal (liquid pressure acting on the air)

or

i.e. We have pressure matching at the interface.

Our body force is

Referring to the Navier-Stokes equation 4.4, we see that our only surviving parts are

It’s important to note that in these problems we have to derive our boundary value conditions! They are not given.

In this discussion, the height was assumed to be constant, with the tangential direction constant and parallel to the surface that the liquid is flowing on. It’s claimed in class that this is actually a consequence of surface tension only! That’s not at all intuitive, but will be covered when we learn about “stability conditions”.

Study note.

Memorizing the NS equation is required for midterm, but more complex stuff (like cylindrical forms of the strain tensor if required) will be given.

Disclaimer.

Problem Q1.

Statement

Solution

We need to express the relation between stress and strain in terms of Young’s modulus and Poisson’s ratio. In terms of Lam\’e parameters our model for the relations between stress and strain for an isotropic solid was given as

Computing the trace

allows us to invert the relationship

In terms of Poisson’s ratio and Young’s modulus , our Lam\’e parameters were found to be

and

Our stress strain model for the relationship for an isotropic solid becomes
we find

or

As a sanity check note that this matches (5.12) of [1], although they use a notation of instead of for Poisson’s ratio. We are now ready to tackle the problem. First we need the trace of the stress tensor

Expanding out the last bits of arithmetic the strain tensor is found to have the form

Note that this is dimensionless, unlike the stress.

Problem Q2.

Statement

Small displacement field in a material is given by

Find

\begin{enumerate}
\item the infinitesimal strain tensor ,
\item the principal strains and the corresponding principal axes at ,
\item Is the body under compression or expansion?
\end{enumerate}

Solution. infinitesimal strain tensor

Diving right in, we have

In matrix form we have

Solution. principle strains and axes

At the point the strain tensor has the value

We wish to diagonalize this, solving the characteristic equation for the eigenvalues

We find the characteristic equation to be

This doesn’t appear to lend itself easily to manual solution (there are no obvious roots to factor out). As expected, since the matrix is symmetric, a plot (\ref{fig:continuumL8:continuumProblemSet1Q2fig1}) shows that all our roots are real

with the corresponding basis (orthonormal eigenvectors), the principle axes are

Solution. Is body under compression or expansion?

To consider this question, suppose that as in the previous part, we determine a basis for which our strain tensor is diagonal with respect to that basis at a given point . We can then simplify the form of the stress tensor at that point in the object

We see that the stress tensor at this point is also necessarily diagonal if the strain is diagonal in that basis (with the implicit assumption here that we are talking about an isotropic material). Noting that the Poisson ratio is bounded according to

so if our trace is positive (as it is in this problem for all points ), then any positive principle strain value will result in a positive stress along that direction). For example at the point of the previous part of this problem (for which ), we have

We see that at this point the and components of stress is positive (expansion in those directions) regardless of the material, and provided that

(i.e. ) the material is under expansion in all directions. For the material at that point is expanding in the and directions, but under compression in the directions.

For a Mathematica notebook that visualizes this part of this problem see https://raw.github.com/peeterjoot/physicsplay/master/notes/phy454/continuumProblemSet1Q2animated.cdf. This animates the stress tensor associated with the problem, for different points and values of Poisson’s ratio , with Mathematica manipulate sliders available to alter these (as well as a zoom control to scale the graphic, keeping the orientation and scale fixed with any variation of the other parameters). This generalizes the solution of the problem (assuming I got it right for the specific point of the problem). The vectors are the orthonormal eigenvectors of the tensor, scaled by the magnitude of the eigenvectors of the stress tensor (also diagonal in the basis of the diagonalized strain tensor at the point in question). For those directions that are under expansive stress, I’ve colored the vectors blue, and for compressive directions, I’ve colored the vectors red.

This requires either a Mathematica client or the free Wolfram CDF player, either of which can run the notebook after it is saved to your computer’s hard drive.

Problem Q3.

Statement

The stress tensor at a point has components given by

Find the traction vector across an area normal to the unit vector

Can you construct a tangent vector on this plane by inspection? What are the components of the force per unit area along the normal and tangent on that surface? (hint: projection of the traction vector.)

Solution

The traction vector, the force per unit volume that holds a body in equilibrium, in coordinate form was

where was the coordinates of the normal to the surface with area . In matrix form, this is just

so our traction vector for this stress tensor and surface normal is just

We also want a vector in the plane, and can pick

or

It’s clear that either of these is normal to (the first can also be computed by normalizing , and the second with one round of Gram-Schmidt). However, neither of these vectors in the plane are particularly interesting since they are completely arbitrary. Let’s instead compute the projection and rejection of the traction vector with respect to the normal. We find for the projection

Our rejection, the component of the traction vector in the plane, is

This gives us a another vector perpendicular to the normal

Wrapping up, we find the decomposition of the traction vector in the direction of the normal and its projection onto the plane to be

The components we can read off by inspection.

Problem Q4.

Statement

The stress tensor of a body is given by

Determine the constant , , and if the body is in equilibrium.

Solution

In the absence of external forces our equilibrium condition was

In matrix form we wish to operate (to the left) with the gradient coordinate vector

So, our conditions for equilibrium will be satisfied when we have

provided , and for integer . If equilibrium is to hold along the plane, then we must either also have or also impose the restriction (for integer ).

A couple other mathematica notebooks

Some of the hand calculations done in this problem set I’ve confirmed using Mathematica. Those notebooks are available here

Disclaimer.

Setup

We got as far as expressing the vector displacement for an isotropic material at a given point in terms of the Lam\’e parameters

P-waves.

Operating on this with the divergence once more, and writing , we have

or

We see that our divergence is governed by a wave equation where the speed of the wave is specified by

so the displacement wave equation is given by

Let’s look at the divergence of the displacement vector in some more detail. By definition this is just

Recall that the strain tensor was defined as

so we have

So the divergence in question can be written in terms of the strain tensor

We also found that the trace of the strain tensor was the relative change in volume. We call this the dilatation. A measure of change in volume as illustrated (badly) in figure (\ref{fig:continuumL7:continuumL7fig1})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL7fig1}
\caption{Illustrating changes in a control volume.}
\end{figure}

This idea can be found nicely animated in the wikipedia page [2].

S-waves.

Now let’s operate on our equation 2.1 with the curl operator

Writing

and observing that (with ), we find

We call this the S-wave equation, and write for the speed of this wave

so that we have

Again, we can find nice animations of this on wikipedia [3].

Relative speeds of the p-waves and s-waves.

Taking ratios of the wave speeds we find

Since both and , we have

Divergence (p-waves) are faster than rotational (s-waves) waves.

In terms of the Poisson ratio , we find

we see that Poisson’s ratio characterizes the speeds of the waves for the medium

Assuming a gradient plus curl representation.

Let’s assume that our displacement can be written in terms of a gradient and curl as we do for the electric field

Inserting this into 2.1 we find

using

Observe that

Here we make use of the fact that an antisymmetric sum of symmetric partials is zero assuming sufficient continuity. Grouping terms we have

When the material is infinite in scope, so that boundary value coupling is not a factor, we can write this as a set of independent P-wave and S-wave equations

Here we have 6 relationships between the components of the strain tensor . Deriving these will be assigned in the homework.

Elastodynamics. Elastic waves.

Reading: Chapter III (section 22 – section 26) of the text [1].

Example: sound or water waves (i.e. waves in a solid or liquid material that comes back to its original position.)

\begin{definition}
\emph{(Elastic Wave)}

An elastic wave is a type of mechanical wave that propagates through or on the surface of a medium. The elasticity of the material provides the restoring force (that returns the material to its original state). The displacement and the restoring force are assumed to be linearly related.
\end{definition}

In symbols we say

and specifically

This is just Newton’s second law, , but expressed in terms of a unit volume.

Should we have an external body force (per unit volume) acting on the body then we must modify this, writing

Note that we are separating out the “original” forces that produced the stress and strain on the object from any constant external forces that act on the body (i.e. a gravitational field).

With

we can expand the stress divergence, for the case of homogeneous deformation, in terms of the Lam\’e parameters

We compute

We find, for homogeneous deformations, that the force per unit volume on our element of mass, in the absence of external forces (the body forces), takes the form

recall that we can decompose our force into components that refer to our direction cosines

Or in tensor form

We call this the traction vector and denote it in vector form as

Constitutive relation.

Reading: section 2, section 4 and section 5 from the text [1].

We can find the relationship between stress and strain, both analytically and experimentally, and call this the Constitutive relation. We prefer to deal with ranges of distortion that are small enough that we can make a linear approximation for this relation. In general such a linear relationship takes the form

Consider the number of components that we are talking about for various rank tensors

We have a lot of components, even for a linear relation between stress and strain. For isotropic materials we model the constitutive relation instead as

For such a modeling of the material the (measured) values and (shear modulus or modulus of rigidity) are called the Lam\’e parameters.

It will be useful to compute the trace of the stress tensor in the form of the constitutive relation for the isotropic model. We find

or

We can now also invert this, to find the trace of the strain tensor in terms of the stress tensor

Substituting back into our original relationship 3.8, and find

which finally provides an inverted expression with the strain tensor expressed in terms of the stress tensor

Special cases.

Hydrostatic compression

Hydrostatic compression is when we have no shear stress, only normal components of the stress matrix is nonzero. Strictly speaking we define Hydrostatic compression as

i.e. not only diagonal, but with all the components of the stress tensor equal.

We can write the trace of the stress tensor as

Now, from our discussion of the strain tensor recall that we found in the limit

allowing us to express the change in volume relative to the original volume in terms of the strain trace

Writing that relative volume difference as we find

or

where is called the Bulk modulus.

Uniaxial stress

Again illustrated in the plane as in figure (\ref{fig:continuumL5:continuumL5fig2})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL5fig2}
\caption{Uniaxial stress.}
\end{figure}

Expanding out 3.12 we have for the element of the strain tensor

or

where is Young’s modulus. Young’s modulus in the text (5.3) is given in terms of the bulk modulus . Using we find

Spherical tensor.

To perform the derivation in spherical coordinates we have some setup to do first, since we need explicit representations of all three unit vectors. The radial vector we can get easily by geometry and find the usual

We can get by geometrical intuition since it the plane unit vector at angle rotated by . That is

A manual derivation.

Doing the calculation pretty much completely with Mathematica is rather unsatisfying. To set up for it let’s first compute the unit vectors from scratch. I’ll use geometric algebra to do this calculation. Consider figure (\ref{fig:qmTwoExamReflection:continuumL2fig5})

We have two sets of rotations, the first is a rotation about the axis by . Writing for the unit bivector in the plane, we rotate

Now we rotate in the plane spanned by and by . With , our vectors in the plane rotate as

(with since ).

Now, these are all the same relations that we could find with coordinate algebra

There’s nothing special in this approach if that is as far as we go, but we can put things in a nice tidy form for computation of the differentials of the unit vectors. Introducing the unit pseudoscalar we can write these in a compact exponential form.

To summarize we have

Taking differentials we find first

Summarizing these differentials we have

A final cleanup is required. While is a vector and has a nicely compact form, we need to decompose this into components in the , and directions. Taking scalar products we have

Summarizing once again, but this time in terms of , and we have

Now we are set to take differentials. With

we have

Squaring this we get the usual spherical polar line scalar line element

With

our differential is

We can add to this and take differences

For each we have

and plugging through that calculation is really all it takes to derive the textbook result. To do this to first order in , we find

We have to consider objects (a control volume) that is small enough that we can consider that we have a point in space limit for the quantities of density and velocity. At the same time we cannot take this limiting process to the extreme, since if we use a control volume that is sufficiently small, quantum and inter-atomic effects would have to be considered.

(ie: is a function of all the initial coordinates, as are the displacements ).

or

with

we have

We write

where we define the \emph{strain tensor} as

Here is a matrix in Cartesian coordinates

We see from 3.11 that is symmetric, so we have

Because any real symmetric matrix can be diagonalized we can write in some coordinate system

If our changes are small enough we can also write approximately, taking the first order term in the square root evaluation

We are also free to define a volume element

So the change of volume is given by the trace

Strain Tensor in cylindrical coordinates.

At the end of the section in the text, the formulas for the spherical and cylindrical versions (to first order) of the strain tensor is given without derivation. Let’s do that derivation for the cylindrical case, which is simpler. It appears that use of explicit vector notation is helpful here, so we write

where

Since and are functions of position, we will need their differentials

but these are just scaled basis vectors

So for our and differentials we find

and

Putting these together we have

For the squared magnitude’s difference from we have

Expanding this out, but dropping all the terms that are quadratic in the components of or its differentials, we have

Grouping all terms, with all the second order terms neglected, we have

From this we can read off the result quoted in the text

Observe that we have to introduce factors of along with all the ‘s, when we factored out the tensor components. That’s an important looking detail, which isn’t obvious unless one works through the derivation.

As with the first order case, we can read off the tensor coordinates by inspection (once we factor out the various factors of and ). The next logical step would be to do the spherical tensor calculation. That would likely be particularily messy if we attempted it in the brute force fashion. Let’s step back and look at the general case, before tackling there sphereical polar form explicitly.

Strain Tensor for general coordinate representation.

Now let’s dispense with the assumption that we have an orthonormal frame. Given an arbitrary, not neccessarily orthonormal, position dependent frame , and its reciprocal frame , as defined by

Our coordinate representation, with summation and dimensionality implied, is

Our differentials are

and

Summing these we have

Taking dot products to form the squares we have

and

Taking the difference we find

To evaluate this, it is useful, albeit messier, to group terms a bit

Here is used to denote summation over the pairs just once, not neccessarily any numeric ordering. For example with , this could be the set .

Cartesian tensor.

In the Cartesian case all the partials of the unit vectors are zero, and we also have no need of upper or lower indexes. We are left with just

However, since we also have , this is

This essentially recovers the result 3.11 derived in class.

Cylindrial tensor.

Now lets do the cylindrical tensor again, but this time without resorting mathematica brute force.

First we recall that all our basis vector derivatives are zero except for the derivatives, and for those we have

We are now set to evaluate the terms in the sum of 3.50 for the cylindrical coordinate system and shouldn’t need Mathematica to do it. Let’s do this one at a time, starting with all the squared differential pairs. Those are, for the value of

For both and all our unit vectors have zero derivatives so we are left respectively with

and

For the term we have

Now, on to the mixed terms. The easiest is the term, for which all the unit vector derivatives are zero, and we are left with just

Now we have the two messy mixed terms. For the , term we get

Finally for the , term we have

To summarize we have, including both first and second order terms,

Factors of have been pulled out so that the portions remaining in the braces are exactly the cylindrical tensor elements as given in the text (except also with the second order terms here). Observe that the pre-calculation of the general formula has allowed an on paper expansion of the cylindrical tensor without too much pain, and this time without requiring Mathematica.

Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

Review. Strain.

Strain is the measure of stretching. This is illustrated pictorially in figure (\ref{fig:continuumL3:continuumL3fig1})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig1}
\caption{Stretched line elements.}
\end{figure}

where is the strain tensor. We found

Why do we have a factor two? Observe that if the deformation is small we can write

so that we find

Suppose for example, that we have a diagonalized strain tensor, then we find

so that

Observe that here again we see this factor of two.

If we have a diagonalized strain tensor, the tensor is of the form

we have

so

Observe that the change in the volume element becomes the trace

How do we use this? Suppose that you are given a strain tensor. This should allow you to compute the stretch in any given direction.

FIXME: find problem and try this.

Stress tensor.

Reading for this section is section 2 from the text associated with the prepared notes [1].

We’d like to consider a macroscopic model that contains the net effects of all the internal forces in the object as depicted in figure (\ref{fig:continuumL3:continuumL3fig2})

We will consider a volume big enough that we won’t have to consider the individual atomic interactions, only the average effects of those interactions. Will will look at the force per unit volume on a differential volume element

The total force on the body is

where is the force per unit volume. We will evaluate this by utilizing the divergence theorem. Recall that this was

We have a small problem, since we have a non-divergence expression of the force here, and it is not immediately obvious that we can apply the divergence theorem. We can deal with this by assuming that we can find a vector valued tensor, so that if we take the divergence of this tensor, we end up with the force. We introduce the quantity

and require this to be a vector. We can then apply the divergence theorem

where is a surface element. We identify this tensor

and

as the force on the surface element . In two dimensions this is illustrated in the following figures (\ref{fig:continuumL3:continuumL3fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig3}
\caption{2D strain tensor.}
\end{figure}

Observe that we use the index above as the direction of the force, and index as the direction normal to the surface.

Note that the strain tensor has the matrix form

We will show later that this tensor is in fact symmetric.

FIXME: given some 3D forces, compute the stress tensor that is associated with it.

Examples of the stress tensor

Example 1. stretch in two opposing directions.

Here, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig4}), the associated (2D) stress tensor takes the simple form

Example 2. stretch in a pair of mutually perpendicular directions

For a pair of perpendicular forces applied in two dimensions, as illustrated in figure (\ref{fig:continuumL3:continuumL3fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{continuumL3fig5}
\caption{Mutually perpendicular forces}
\end{figure}

our stress tensor now just takes the form

It’s easy to imagine now how to get some more general stress tensors, should we make a change of basis that rotates our frame.

Example 3. radial stretch

Suppose we have a fire fighter’s safety net, used to catch somebody jumping from a burning building (do they ever do that outside of movies?), as in figure (\ref{fig:continuumL3:continuumL3fig6}). Each of the firefighters contributes to the stretch.